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问题:
Given is an array of bodies that interact in some way with each other. As a newbie I approached it as I would do it in some other language:
struct Body {
x: i16,
y: i16,
v: i16,
}
fn main() {
let mut bodies = Vec::<Body>::new();
bodies.push(Body { x: 10, y: 10, v: 0 });
bodies.push(Body { x: 20, y: 30, v: 0 });
// keep it simple and loop only twice
for i in 0..2 {
println!("Turn {}", i);
for b_outer in bodies.iter() {
println!("x:{}, y:{}, v:{}", b_outer.x, b_outer.y, b_outer.v);
let mut a = b_outer.v;
for b_inner in bodies.iter() {
// for simplicity I ignore here to continue in case b_outer == b_inner
// just do some calculation
a = a + b_outer.x * b_inner.x;
println!(
" x:{}, y:{}, v:{}, a:{}",
b_inner.x,
b_inner.y,
b_inner.v,
a
);
}
// updating b_outer.v fails
b_outer.v = a;
}
}
}
Updating of b_outer.v
after the inner loop has finished fails:
error[E0594]: cannot assign to immutable field `b_outer.v`
--> src/main.rs:32:13
|
32 | b_outer.v = a;
| ^^^^^^^^^^^^^ cannot mutably borrow immutable field
Making b_outer
mutable:
for b_outer in bodies.iter_mut() { ...
doesn't work either:
error[E0502]: cannot borrow `bodies` as mutable because it is also borrowed as immutable
--> src/main.rs:19:32
|
16 | for b_outer in bodies.iter() {
| ------ immutable borrow occurs here
...
19 | for b_inner in bodies.iter_mut() {
| ^^^^^^ mutable borrow occurs here
...
33 | }
| - immutable borrow ends here
Now I'm stuck. What's the Rust approach to update b_outer.v
after the inner loop has finished?
回答1:
For what it's worth, I think the error message is telling you that your code has a logic problem. If you update the vector between iterations of the inner loop, then those changes will be used for subsequent iterations. Let's look at a smaller example where we compute the windowed-average of an array item and its neighbors:
[2, 0, 2, 0, 2] // input
[2/3, 4/3, 2/3, 4/3, 2/3] // expected output (out-of-bounds counts as 0)
[2/3, 0, 2, 0, 2] // input after round 1
[2/3, 8/9, 2, 0, 2] // input after round 2
[2/3, 8/9, 26/9, 0, 2] // input after round 3
// I got bored here
I'd suggest computing the output into a temporary array and then swap them:
#[derive(Debug)]
struct Body {
x: i16,
y: i16,
v: i16,
}
fn main() {
let mut bodies = vec![
Body { x: 10, y: 10, v: 0 },
Body { x: 20, y: 30, v: 0 },
];
for _ in 0..2 {
let next_bodies = bodies
.iter()
.map(|b| {
let next_v = bodies
.iter()
.fold(b.v, { |a, b_inner| a + b.x * b_inner.x });
Body { v: next_v, ..*b }
})
.collect();
bodies = next_bodies;
}
println!("{:?}", bodies);
}
Output:
[Body { x: 10, y: 10, v: 600 }, Body { x: 20, y: 30, v: 1200 }]
If you really concerned about memory performance, you could create a total of two vectors, size them appropriately, then alternate between the two. The code would be uglier though.
As Matthieu M. said, you could use Cell
or RefCell
, which both grant you inner mutability:
use std::cell::Cell;
#[derive(Debug, Copy, Clone)]
struct Body {
x: i16,
y: i16,
v: i16,
}
fn main() {
let bodies = vec![
Cell::new(Body { x: 10, y: 10, v: 0 }),
Cell::new(Body { x: 20, y: 30, v: 0 }),
];
for i in 0..2 {
println!("Turn {}", i);
for b_outer_cell in &bodies {
let mut b_outer = b_outer_cell.get();
println!("{:?}", b_outer);
let mut a = b_outer.v;
for b_inner in &bodies {
let b_inner = b_inner.get();
a = a + b_outer.x * b_inner.x;
println!("{:?}, a: {}", b_inner, a);
}
b_outer.v = a;
b_outer_cell.set(b_outer);
}
}
println!("{:?}", bodies);
}
[Cell { value: Body { x: 10, y: 10, v: 600 } }, Cell { value: Body { x: 20, y: 30, v: 1200 } }]
回答2:
I know the question is like 2 years old, but I got curious about it.
This C# program produces the original desired output:
var bodies = new[] { new Body { X = 10, Y = 10, V = 0 },
new Body { X = 20, Y = 30, V = 0 } };
for (int i = 0; i < 2; i++)
{
Console.WriteLine("Turn {0}", i);
foreach (var bOuter in bodies)
{
Console.WriteLine("x:{0}, y:{1}, v:{2}", bOuter.X, bOuter.Y, bOuter.V);
var a = bOuter.V;
foreach (var bInner in bodies)
{
a = a + bOuter.X * bInner.X;
Console.WriteLine(" x:{0}, y:{1}, v:{2}, a:{3}", bInner.X, bInner.Y, bInner.V, a);
}
bOuter.V = a;
}
}
Since only v
is ever changed, we could change the struct to something like this:
struct Body {
x: i16,
y: i16,
v: Cell<i16>,
}
Now I'm able to mutate v
, and the program becomes:
// keep it simple and loop only twice
for i in 0..2 {
println!("Turn {}", i);
for b_outer in bodies.iter() {
let mut a = b_outer.v.get();
println!("x:{}, y:{}, v:{}", b_outer.x, b_outer.y, a);
for b_inner in bodies.iter() {
a = a + (b_outer.x * b_inner.x);
println!(
" x:{}, y:{}, v:{}, a:{}",
b_inner.x,
b_inner.y,
b_inner.v.get(),
a
);
}
b_outer.v.set(a);
}
}
It produces the same output as the C# program above. The "downside" is that whenever you want to work with v
, you need use get()
or into_inner()
. There may be other downsides I'm not aware of.
回答3:
I decided to split the structure in one that is used as a base for the calculation (input) in the inner loop (b_inner
) and one that gathers the results (output). After the inner loop finished, the input structure is updated in the outer loop (b_outer
) and calculation starts with the next body.
What's now not so nice that I have to deal with two structures and you don't see their relation from the declaration.
#[derive(Debug)]
struct Body {
x: i16,
y: i16,
}
struct Velocity {
vx: i16,
}
fn main() {
let mut bodies = Vec::<Body>::new();
let mut velocities = Vec::<Velocity>::new();
bodies.push(Body { x: 10, y: 10 });
bodies.push(Body { x: 20, y: 30 });
velocities.push(Velocity { vx: 0 });
velocities.push(Velocity { vx: 0 });
// keep it simple and loop only twice
for i in 0..2 {
println!("Turn {}", i);
for (i, b_outer) in bodies.iter().enumerate() {
println!("x:{}, y:{}, v:{}", b_outer.x, b_outer.y, velocities[i].vx);
let v = velocities.get_mut(i).unwrap();
let mut a = v.vx;
for b_inner in bodies.iter() {
// for simplicity I ignore here to continue in case b_outer == b_inner
// just do some calculation
a = a + b_outer.x * b_inner.x;
println!(" x:{}, y:{}, v:{}, a:{}", b_inner.x, b_inner.y, v.vx, a);
}
v.vx = a;
}
}
println!("{:?}", bodies);
}
Output:
[Body { x: 10, y: 10 }, Body { x: 20, y: 30 }]